Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x-3y &= 3 \\ 2x-2y &= -7\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-2y = -2x-7$ Divide both sides by $-2$ to isolate $y$ $y = {x + \dfrac{7}{2}}$ Substitute this expression for $y$ in the first equation. $9x-3({x + \dfrac{7}{2}}) = 3$ $9x - 3x - \dfrac{21}{2} = 3$ Simplify by combining terms, then solve for $x$ $6x - \dfrac{21}{2} = 3$ $6x = \dfrac{27}{2}$ $x = \dfrac{9}{4}$ Substitute $\dfrac{9}{4}$ for $x$ back into the top equation. $9( \dfrac{9}{4})-3y = 3$ $\dfrac{81}{4}-3y = 3$ $-3y = -\dfrac{69}{4}$ $y = \dfrac{23}{4}$ The solution is $\enspace x = \dfrac{9}{4}, \enspace y = \dfrac{23}{4}$.